Complex Numbers | 复数考点精讲

📚 Complex Numbers | 复数考点精讲

Complex numbers are a fascinating extension of the real number system and a key topic in AQA Level 2 Certificate in Further Mathematics. They allow us to solve equations that have no real solutions, such as x² + 1 = 0, and open up a powerful new world of algebra and geometry. In this article, we will walk through every essential concept — from the imaginary unit i to Argand diagrams — with clear explanations and practical examples tailored to the AQA specification. Whether you are meeting complex numbers for the first time or revising for your exam, this guide will help you master the topic with confidence.

复数是实数系统的一个迷人扩展，也是 AQA Level 2 进阶数学证书中的重要考点。它们让我们能够求解没有实数解的方程，比如 x² + 1 = 0，并开启一个强大的代数与几何新世界。在本文中，我们将逐一讲解每一个核心概念——从虚数单位 i 到阿 Argand 图——结合清晰易懂的说明和贴近 AQA 考纲的实用例题。无论你是初次接触复数，还是在为考试复习，本指南都将帮助你自信掌握这一主题。

1. Introduction to Complex Numbers | 复数概述

Complex numbers were invented to provide solutions to equations that cannot be solved using only real numbers. For instance, the equation x² = –1 has no real solution because squaring any real number gives a non-negative result. To overcome this, mathematicians introduced a new type of number, called an imaginary number, and from it built the complex number system. A complex number is any number that can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit with the property i² = –1.

复数的发明是为了给仅用实数无法求解的方程提供解。例如，方程 x² = –1 没有实数解，因为任何实数的平方都是非负的。为了克服这一点，数学家引入了一种新的数——虚数，并由此构建了复数系统。复数是指可以写成 a + bi 形式的任何数，其中 a 和 b 是实数，i 是虚数单位，满足性质 i² = –1。

In a complex number a + bi, a is called the real part and b is called the imaginary part. For example, in 3 + 4i, the real part is 3 and the imaginary part is 4. If b = 0, the number is purely real; if a = 0, the number is purely imaginary. This structure merges the real and imaginary worlds into a single coherent framework.

在复数 a + bi 中，a 称为实部，b 称为虚部。例如，在 3 + 4i 中，实部是 3，虚部是 4。如果 b = 0，这个数就是纯实数；如果 a = 0，它就是纯虚数。这一结构将实数和虚数世界融合成了一个统一的体系。

2. The Imaginary Unit i | 虚数单位 i

The core of complex numbers is the imaginary unit i, defined by the equation i² = –1. From this definition, we immediately see that i is not a real number. It behaves like a mathematical ‘building block’ for all other imaginary and complex numbers. The square root of a negative number can now be expressed using i: for any positive real number p, √(–p) = i√p.

复数的核心是虚数单位 i，其定义为 i² = –1。从这一定义我们立即看出 i 不是实数。它就像一个数学“积木”，可以用来构造所有其他虚数和复数。现在，任何负数的平方根都可以用 i 来表示：对于任意正实数 p，√(–p) = i√p。

It is vital to remember that i is not a variable but a specific mathematical constant. Treat it just like you would treat √5 or π — it has a fixed meaning. When simplifying expressions, always replace i² with –1 as soon as it appears.

务必记住，i 不是变量，而是一个特定的数学常数。就像处理 √5 或 π 一样去处理它——它具有固定的含义。在进行表达式化简时，一旦出现 i²，就立即用 –1 替换。

3. Powers of i | i 的幂

The powers of i follow a cyclic pattern of period 4. Starting from i¹ = i, i² = –1, i³ = i²·i = –i, i⁴ = (i²)² = 1, and then the cycle repeats: i⁵ = i, i⁶ = –1, i⁷ = –i, i⁸ = 1, and so on. This means any power of i can be reduced by dividing the exponent by 4 and taking the remainder.

i 的幂遵循一个周期为 4 的循环模式。从 i¹ = i 开始，i² = –1，i³ = i²·i = –i，i⁴ = (i²)² = 1，然后循环重复：i⁵ = i，i⁶ = –1，i⁷ = –i，i⁸ = 1，以此类推。这意味着任何 i 的幂都可以通过用指数除以 4 取余数来化简。

For example, to simplify i²³, divide 23 by 4: 23 = 5×4 + 3, so i²³ = i³ = –i. This trick is extremely useful in simplifying products and quotients of complex numbers, and it frequently appears in AQA exam questions. Always remember: remainder 0 → 1, 1 → i, 2 → –1, 3 → –i.

例如，要化简 i²³，将 23 除以 4：23 = 5×4 + 3，所以 i²³ = i³ = –i。这个技巧在化简复数的乘积和商时极为有用，并且经常出现在 AQA 的考题中。始终牢记：余数 0 → 1，余数 1 → i，余数 2 → –1，余数 3 → –i。

4. Complex Number Form a + bi | 复数形式 a + bi

Every complex number can be written uniquely as a + bi, where a, b ∈ ℝ. This is known as Cartesian form. The real part a and the imaginary part b are simply real numbers. When writing complex numbers, we usually put the i after the coefficient, e.g., 2 + 5i, not 2 + i5. Purely real numbers are written without i (e.g., 7 = 7 + 0i), and purely imaginary numbers appear as 0 + bi or simply bi (e.g., –3i).

每一个复数都可以唯一地写成 a + bi 的形式，其中 a, b ∈ ℝ。这被称为笛卡儿形式。实部 a 和虚部 b 只是普通的实数。在书写复数时，我们通常把 i 放在系数之后，例如 2 + 5i，而不是 2 + i5。纯实数就写成没有 i 的形式（如 7 = 7 + 0i），纯虚数则写为 0 + bi 或直接写 bi（如 –3i）。

This standard form makes addition and multiplication straightforward because we can treat i almost like an algebraic variable, with the crucial extra rule that i² = –1. When simplifying, always collect the real and imaginary parts separately. For instance, the sum (3 + 2i) + (1 – 5i) = (3+1) + (2–5)i = 4 – 3i.

这种标准形式使得加法和乘法变得简单，因为我们可以把 i 几乎当作代数变量来处理，再加上关键的一条额外规则 i² = –1。在进行化简时，始终将实部和虚部分别合并。例如，和 (3 + 2i) + (1 – 5i) = (3+1) + (2–5)i = 4 – 3i。

5. Adding and Subtracting Complex Numbers | 复数的加减

Addition and subtraction of complex numbers are carried out by combining the real parts and the imaginary parts separately. If z₁ = a + bi and z₂ = c + di, then z₁ + z₂ = (a + c) + (b + d)i, and z₁ – z₂ = (a – c) + (b – d)i. These operations are both commutative and associative, just as with real numbers.

复数的加减法通过分别合并实部和虚部来完成。如果 z₁ = a + bi，z₂ = c + di，那么 z₁ + z₂ = (a + c) + (b + d)i，而 z₁ – z₂ = (a – c) + (b – d)i。这些运算满足交换律和结合律，就像实数运算一样。

Be careful with signs when subtracting: the minus sign applies to both the real and imaginary parts of the second number. For example, (5 + 3i) – (2 – 4i) = (5 – 2) + (3 – (–4))i = 3 + 7i. Visualising complex addition on an Argand diagram shows it is equivalent to vector addition.

做减法时要小心符号：减号同时作用于第二个数的实部和虚部。例如，(5 + 3i) – (2 – 4i) = (5 – 2) + (3 – (–4))i = 3 + 7i。在阿 Argand 图上将复数加法可视化，可以看到它等价于向量的加法。

6. Multiplying Complex Numbers | 复数乘法

Multiplication of complex numbers uses the same distributive law as algebra, together with the fact that i² = –1. For z₁ = a + bi and z₂ = c + di, the product is:

(a + bi)(c + di) = ac + adi + bci + bdi²

= (ac – bd) + (ad + bc)i.

复数的乘法使用与代数相同的分配律，并结合 i² = –1 这一事实。对于 z₁ = a + bi 和 z₂ = c + di，其乘积为：

(a + bi)(c + di) = ac + adi + bci + bdi²

= (ac – bd) + (ad + bc)i。

You can either memorise this formula or simply expand brackets and simplify. For example, compute (2 + 3i)(1 – 4i):

= 2(1 – 4i) + 3i(1 – 4i) = 2 – 8i + 3i – 12i²

= 2 – 5i – 12(–1) = 2 – 5i + 12 = 14 – 5i.

你可以记住这个公式，也可以直接展开括号并化简。例如，计算 (2 + 3i)(1 – 4i)：

= 2(1 – 4i) + 3i(1 – 4i) = 2 – 8i + 3i – 12i²

= 2 – 5i – 12(–1) = 2 – 5i + 12 = 14 – 5i。

When multiplying purely imaginary numbers, watch the i²: e.g., (3i)(4i) = 12i² = –12, which is real. This is a common pitfall in exams — never forget to replace i² with –1.

当两个纯虚数相乘时，要留意 i²：例如 (3i)(4i) = 12i² = –12，这是一个实数。这在考试中是一个常见的陷阱——永远不要忘记用 –1 替换 i²。

7. Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is denoted by z* or sometimes z̄, and is defined as a – bi. It is obtained by simply changing the sign of the imaginary part. Conjugates are extremely useful because the product of a complex number and its conjugate is always a non-negative real number: (a + bi)(a – bi) = a² – (bi)² = a² + b².

复数 z = a + bi 的共轭记为 z*（有时也写作 z̄），定义为 a – bi。它只需将虚部的符号改变即可得到。共轭复数极其有用，因为一个复数与其共轭的乘积总是一个非负实数：(a + bi)(a – bi) = a² – (bi)² = a² + b²。

This property is the key to dividing complex numbers and to finding the modulus of a complex number. The sum and difference of a complex number and its conjugate also give simple real or imaginary results: z + z* = 2a (real), and z – z* = 2bi (purely imaginary).

这一性质是复数除法和求复数模的关键。一个复数与其共轭的和与差也会得到简单的实数或纯虚数结果：z + z* = 2a（实数），z – z* = 2bi（纯虚数）。

8. Dividing Complex Numbers | 复数除法

To divide one complex number by another, multiply both the numerator and the denominator by the complex conjugate of the denominator. This turns the denominator into a real number, allowing us to separate the result into real and imaginary parts. For z₁ = a + bi and z₂ = c + di, the quotient is:

(a + bi)/(c + di) = (a + bi)(c – di) / (c + di)(c – di) = [(ac + bd) + (bc – ad)i] / (c² + d²).

要将一个复数除以另一个复数，将分子和分母同时乘以分母的共轭复数。这样就将分母变成了实数，从而我们可以将结果分离为实部和虚部。对于 z₁ = a + bi 和 z₂ = c + di，其商为：

(a + bi)/(c + di) = (a + bi)(c – di) / (c + di)(c – di) = [(ac + bd) + (bc – ad)i] / (c² + d²)。

A typical exam-style example: simplify (3 + 2i)/(1 – i). Multiply top and bottom by the conjugate of the denominator, 1 + i:

= (3 + 2i)(1 + i) / (1 – i)(1 + i)

= (3 + 3i + 2i + 2i²) / (1 – i²)

= (3 + 5i – 2) / (1 + 1) = (1 + 5i)/2 = 0.5 + 2.5i.

一个典型的考试题示例：化简 (3 + 2i)/(1 – i)。将分子分母同时乘以分母的共轭 1 + i：

= (3 + 2i)(1 + i) / (1 – i)(1 + i)

= (3 + 3i + 2i + 2i²) / (1 – i²)

= (3 + 5i – 2) / (1 + 1) = (1 + 5i)/2 = 0.5 + 2.5i。

9. Solving Quadratic Equations with Complex Roots | 解有复数根的二次方程

One of the most important applications of complex numbers is solving quadratic equations where the discriminant is negative. For an equation ax² + bx + c = 0, if b² – 4ac < 0, the roots are a pair of complex conjugates given by the quadratic formula:
x = [–b ± i√(4ac – b²)] / (2a).

复数最重要的应用之一是求解判别式为负的二次方程。对于方程 ax² + bx + c = 0，如果 b² – 4ac < 0，那么它的两根是一对共轭复数，由求根公式给出：
x = [–b ± i√(4ac – b²)] / (2a)。

For example, solve x² + 4x + 13 = 0.

Discriminant Δ = 4² – 4·1·13 = 16 – 52 = –36.

√Δ = √(–36) = 6i.

So x = (–4 ± 6i)/2 = –2 ± 3i.

The two roots are –2 + 3i and –2 – 3i, which are complex conjugates. This always happens when coefficients are real.

例如，求解 x² + 4x + 13 = 0。

判别式 Δ = 4² – 4·1·13 = 16 – 52 = –36。

√Δ = √(–36) = 6i。

因此 x = (–4 ± 6i)/2 = –2 ± 3i。

两个根为 –2 + 3i 和 –2 – 3i，它们互为共轭复数。当系数为实数时，这种情况总是出现。

10. Argand Diagrams (Basics) | 阿 Argand 图基础

An Argand diagram is a coordinate plane used to represent complex numbers graphically. The horizontal axis (x-axis) represents the real part, and the vertical axis (y-axis) represents the imaginary part. The complex number z = a + bi corresponds to the point (a, b) or the position vector from the origin to (a, b).

阿 Argand 图是一个用于图形化表示复数的坐标平面。横轴（x 轴）代表实部，纵轴（y 轴）代表虚部。复数 z = a + bi 对应于点 (a, b) 或从原点到 (a, b) 的位置向量。

Addition of complex numbers on an Argand diagram follows the parallelogram rule for vectors. The conjugate z* = a – bi is simply the reflection of z in the real axis. This visual approach helps to understand the geometry of complex numbers, though the AQA GCSE Further Maths exam mainly tests the algebraic side.

在阿 Argand 图上，复数的加法遵循向量的平行四边形法则。共轭复数 z* = a – bi 就是 z 关于实轴的反射。这种直观的方法有助于理解复数的几何意义，不过 AQA GCSE 进阶数学考试主要考查的是代数方面。

11. Modulus and Argument (Introduction) | 模与辐角介绍

Although not always required in depth at GCSE, it is helpful to know that the modulus of a complex number z = a + bi, written |z|, is its distance from the origin on the Argand diagram. It is given by |z| = √(a² + b²). The argument of z, arg(z), is the angle the vector makes with the positive real axis, usually measured in radians or degrees. AQA Level 2 Further Maths may ask for the modulus but rarely the argument in a complex calculation question.

虽然在 GCSE 阶段不总是要求深入掌握，但了解一下还是有帮助的：复数 z = a + bi 的模，记作 |z|，是它在阿 Argand 图上到原点的距离。由 |z| = √(a² + b²) 给出。复数的辐角 arg(z) 是该向量与正实轴之间的夹角，通常以弧度或度来度量。AQA Level 2 进阶数学可能会要求计算模，但在复杂计算题中几乎不涉及辐角。

The modulus connects directly to the conjugate product: z z* = |z|² = a² + b². This relationship is frequently used when dividing complex numbers or proving identities. For example, |3 + 4i| = √(3² + 4²) = 5.

模与共轭乘积直接相关：z z* = |z|² = a² + b²。这个关系在进行复数除法或证明恒等式时经常用到。例如，|3 + 4i| = √(3² + 4²) = 5。

12. Summary and Exam Tips | 总结与考试技巧

To excel in AQA Level 2 Further Maths questions on complex numbers, keep these key points in mind:

Always express answers in the form a + bi.
Simplify i² to –1 immediately.
For division, multiply numerator and denominator by the conjugate of the denominator.
When solving quadratics, if the discriminant is negative, rewrite √(negative) as i√(positive).
Remember the cyclic powers of i — divide exponent by 4 and use remainder.
Check that complex roots of real-coefficient equations appear in conjugate pairs.

要在 AQA Level 2 进阶数学复数题目中取得优异成绩，请牢记以下要点：

始终将结果写成 a + bi 的形式。
立即将 i² 化为 –1。
做除法时，将分子分母同乘以分母的共轭。
解二次方程时，若判别式为负，将 √(负数) 重写为 i√(正数)。
牢记 i 的幂的循环规律——将指数除以 4，用余数计算。
检查实系数方程的复数根是否成对出现为共轭复数。

Practice is vital. Work through past paper questions, especially those involving algebraic manipulation of complex expressions and solving quadratic equations. The good news is that complex number questions are often very structured, so methodical work earns method marks.

练习至关重要。多做往年真题，尤其是那些涉及复数代数运算和求解二次方程的题目。好消息是，复数题目通常结构清晰，按步骤解题就能拿到步骤分。

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Complex Numbers | 复数考点精讲